$\int_{-2}^{2}|1 - x^2|dx =$
A4
B2
C-2
D0
Answer & Solution
Correct answer: A. 4
$|1-x^2|$ changes sign at $x=\pm 1$. Also, the integrand is even, so
$$\int_{-2}^{2}|1-x^2|\,dx=2\int_{0}^{2}|1-x^2|\,dx.$$
Now split at $x=1$:
$$2\int_{0}^{2}|1-x^2|\,dx=2\left(\int_{0}^{1}(1-x^2)\,dx+\int_{1}^{2}(x^2-1)\,dx\right).$$
Evaluate each part:
$$\int_{0}^{1}(1-x^2)\,dx=\left[x-\frac{x^3}{3}\right]_{0}^{1}=\frac{2}{3}.$$
$$\int_{1}^{2}(x^2-1)\,dx=\left[\frac{x^3}{3}-x\right]_{1}^{2}=\frac{4}{3}.$$
So
$$2\left(\frac{2}{3}+\frac{4}{3}\right)=2\cdot 2=4.$$
Comparing with the options, the correct choice is $4$, which is option $A$.
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